Question: A certain circle can be represented by the following equation. $x^2+y^2-18x+20y+177=0$ What is the center of this circle ? $($
Explanation: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2-18x+20y+177&=0\\\\ x^2+y^2-18x+20y&=-177\\\\ (x^2-18x)+(y^2+20y)&=-177 \text{(rearrange terms)}\\\\ (x^2-18x{+81})+(y^2+20y{+100})&=-177{+81}{+100}\end{aligned}$ Notice that we must add ${81}$ and ${100}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 81 and 100?] Writing the equation in standard form $\begin{aligned}(x^2-18x{+81})+(y^2+20y{+100})&=-177{+81}{+100}\\\\ (x-9)^2+(y+{10})^2&=4\\\\ (x-9)^2+(y-(-10))^2&=2^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(9,-10)$ and has a radius of $2$ units. Summary The circle is centered at $(9,-10)$. The circle has a radius of $2$ units.